8x-3 ≥ -3x^2
3x^2 + 8x - 3 ≥ 0
(x + 3)(3x - 1) ≥ 0
critical values are x = -3 and x = 1/3
visualize the parabola y = 3x^2 + 8x - 3
it would have x-intercepts of -3 and 1/3 and be above the x-axis , y≥0 , for x ≤ -3 or x ≥ 1/3
so :
x ≤ -3 or x ≥ 1/3
Solve the inequality algebraically
8x-3 ≥ -3x^2
1 answer